solve 4(x)^2 -19x - 5 =0

The purpose of this question is find values of x when y = 0 in the graph (ie when the curve crosses the x axis) There are 2 steps for this: 1) factorise the quadratics equation (2) solve for x Step 1 4(x)^2 - 20x + 1x -5 (break down -19 into 2 numbers - very important concept) 4x(x-5) + 1(x-5) (pull out the common factor of x) (4x + 1)(x - 5) (pull out the second factor of x ) Step 2 find the values of x using the 2 factors 1) 4x+1 = 0 --> 4x = -1 --> x = -1/4 2) x-5 = 0 --> x = 5 Therefore, the graph 4(x)^2 -19x - 5 crosses x axis when x = -1/4 and 5